当前位置：文档之家 › 2005 AMC 12A Problems and Solution

2005 AMC 12A Problems and Solution

  • 格式：doc
  • 大小：935.50 KB
  • 文档页数：19

2005 AMC 12A Problems and Solution Problem 1Two is of and of . What is ?Solution. Problem 2The equations and have the sam e solution. What is the value of ?SolutionProblem 3A rectangle with diagonal length is twice as long as it is wide. What is the area of the rectangle?SolutionLet be the width, so the length is . By the Pythagorean Theorem,. The area of the rectangle is.Problem 4A store norm ally sells windows at $each. This week the store is offering one free window for each purchase of four. Dave needs seven windows and Doug needs eight windows. How much will they save if they purchase the windows together rather than separately?SolutionFor windows, the store offers a discount of (floor function). Davereceives a discount of and Doug receives a disc ount of. These am ount to dollars in discounts. Together, they receivea discount of , so they save .Problem 5The average (m ean) of 20 numbers is 30, and the average of 30 other num bers is 20. What is the average of all 50 numbers?SolutionThe sum of the first 20 numbers is and the sum of the other 30 numbers is. Hence the overall average is .Problem 6Josh and Mike live 13 miles apart. Yesterday, Josh started to ride his bicycle toward Mike's house. A little later Mike started to ride his bicycle toward Josh's house. Whenthey m et, Josh had ridden for twice the length of tim e as Mike and at four-fifths of Mike's rate. How m any miles had Mike ridden when they m et?SolutionLet be the distances traveled by Josh and Mike, respectively, and let bethe tim e and rate of Mike. Using , we have that and. Then.Problem 7Square is inside the square so that each side of can beextended to pass through a vertex of . Square has side lengthand . What is the area of the inner square ?SolutionArguable the hardest part of this question is to visualize the diagram. Since each side of can be extended to pass through a vertex of , we realizethat must be tilted in such a fashion. Let a side of be .Notice the right triangle (in blue) with legs and hypotenuse. By the Pythagorean Theorem, we have. Thus,Problem 8Let , and be digits withWhat is ?SolutionClearly the two quantities are both integers, so we check the prime factorization of. It is easy to see now that works, so theanswer is .Problem 9There are two values of for which the equation has only one solution for . What is the sum of these values of ?SolutionSolution 1A quadratic equation always has two roots, unless it has a double root. That m eans we can write the quadratic as a square, and the coefficients 4 and 9 suggest this.Completing the square, , so. The sum of these is .Solution 2Another method would be to use the quadratic form ula, since our coefficient isgiven as 4, the coefficient is and the constant term is . Hence,Because we want only a single solution for, the determinant m ust equal 0. Therefore, we can write whichfactors to ; using Vieta's formulas we see that the sum of thesolutions for is the opposite of the coeffi cient of , or .Using the discriminant, the result m ust equal .Therefore, or , giving a sum of .Problem 10A wooden cube units on a side is painted red on all six faces and then cut into unit cubes. Exactly one-fourth of the total number of faces of the unit cubes are red. What is ?SolutionThere are sides total on the unit cubes, and are painted red.Problem 11How many three-digit numbers satisfy the property that the m iddle digit is the average of the first and the last digits?SolutionSolution 1Let the digits be so that . In order for this to be an integer,and have to have the sam e parity. There are possibilities for , and for .depends on the value of both and and is unique for each . Thus ouranswer is .Thus, the three digits form an arithmetic sequence.▪If the num bers are all the sam e, then there are possible three-digit numbers.▪If the numbers are different, then we count the number of strictly increasingarithmetic sequences between and and multiply by 2 for the decreasing ones:This gives us . However, the question asks for three-digit numbers, so we have to ignore the four sequences starting with . Thus our answeris .Problem 12A line passes through and . How m any other points withinteger coordinates are on the line and strictly between and ?SolutionFor convenience’s sake, we can transform to the origin and to (thisdoes not change the problem). The line has the equation.The coordinates are integers if , so the values of are , with a totalof coordinates.Problem 13The regular 5-point star is drawn and in each vertex, there is a number.Each and are chosen such that all 5 of them cam e from set. Each letter is a different number (so one possible way is). Let be the sum of the num bers onand , and so forth. If and form an arithmetic sequence(not necessarily in increasing order), find the value of .Solution. The sumwill always be , so the arithm etic sequence has a sum of . Since is the middle term, it must be theaverage of the five num bers, of .Problem 14On a standard die one of the dots is removed at random with each dot equally likely to be chosen. The die is then rolled. What is the probability that the top face has an odd number of dots?SolutionThere are dots total. Casework:▪The dot is rem oved from an even face. There is a chance ofthis happening. Then there are 4 odd faces, giving us a probability of .▪The dot is rem oved from an odd face. There is a chance ofthis happening. Then there are 2 odd faces, giving us a probability of .Thus the answer is .Problem 15Let be a diam eter of a circle and be a point on with . Letand be points on the circle such that and is a second diameter.What is the ratio of the area of to the area of ?SolutionSolution 1Notice that the bases of both triangles are diameters of the circle. Hence the ratio ofthe areas is just the ratio of the heights of the triangles, or (is the foot of the perpendicular from to ).Call the radius . Then , . Using the PythagoreanTheorem in , we get .Now we have to find . Notice , so we can write theproportion:By the Pythagorean Theorem in , we have.Our answer is .Solution 2Let the centre of the circle be .Note that .is midpoint of .is midpoint of Area of Area of Area ofArea of .Problem 16Three circles of radius are drawn in the first quadrant of the -plane. The first circle is tangent to both axes, the second is tangent to the first circle and the -axis, and the third is tangent to the first circle and the -axis. A circle of radius istangent to both axes and to the second and third circles. What is ?SolutionWithout loss of generality, let . Draw the segm ent between the center of thethird circle and the large circle; this has length . We then draw the radius of the large circle that is perpendicular to the x-axis, and draw the perpendicular from this radius to the center of the third circle. This gives us a right triangle with legsand hypotenuse. The Pythagorean Theorem yields:Quite obviously , so and .Problem 17A unit cube is cut twice to form three triangular prisms, two of which are congruent, as shown in Figure 1. The cube is then cut in the sam e m anner along the dashed lines shown in Figure 2. This creates nine pieces. What is the volum e of the piece that contains vertex ?SolutionIt is a pyramid, so .Problem 18Call a number "prime-looking" if it is composite but not divisible by 2, 3, or 5. The three sm allest prime-looking numbers are 49, 77, and 91. There are 168 prime numbers less than 1000. How many prime-looking numbers are there less than 1000?SolutionThe given states that there are prime numbers less than , which is a fact we m ust som ehow utilize. Since there seem s to be no easy way to directly calculate the number of "prime-looking" numbers, we can apply com plementary counting.We can split the num bers from to into several groups:. Hence, the num ber of prime-looking numbers is(note that are primes).We can calculate using the Principle of Inclusion-Exclusion: (thevalues of and their intersections can be found quite easily)Substituting, we find that our answer is .Problem 19A faulty car odom eter proceeds from digit 3 to digit 5, always skipping the digit 4, regardless of position. If the odometer now reads 002005, how many miles has the car actually traveled?SolutionWe find the number of numbers with a and subtract from. Quick counting tells us that there are numbers with a 4 in the hundreds place, numberswith a 4 in the tens place, and numbers with a 4 in the units place (counting). Now we apply the Principle of Inclusion-Exclusion. There are numberswith a 4 in the hundreds and in the tens, and for both the other two intersections.The intersection of all three sets is just . So we get:Alternatively, consider that counting without the number is almost equivalent tocounting in base ; only, in base , the number is not counted. Since is skipped,the sym bol represents miles of travel, and we have traveled miles. Bybasic conversion, .Problem 20For each in , defineLet , and for each integer . Forhow m any values of in is ?SolutionFor the two functions and ,wecan see that as long as is between and , will be in the right domain.Therefore, we don't need to worry about the dom ain of . Also, every tim e wechange , the final equation will be in a different form and thus we will get adifferent value of x. Every tim e we have two choices for ) and altogether wehave to choose tim es. Thus, .Problem 21How many ordered triples of integers, with , , and ,satisfy both and ?SolutionCasework upon :▪: Then . Thus we get .▪: Then . Thus we get .▪: Then the exponent of becom es huge, and since there is noway we can satisfy the second condition. Hence we have two ordered triples . Problem 22A rectangular box is inscribed in a sphere of radius . The surface area of is 384,and the sum of the lengths of its 12 edges is 112. What is ?SolutionThe box P has dimensions , , and . Therefore,▪▪Now we m ake a formula for . Since the diam eter of the sphere is the spacediagonal of the box,▪We square :▪We get thatProblem 23Two distinct num bers and are chosen randomly from the set .What is the probability that is an integer?SolutionLet , so . Define , ; then , so .Here we can just make a table and count the num ber of values of per value of .The largest possible value of is 12, and we get.The total number of ways to pick two distinct numbers is ,so we get a probability of .Problem 24Let . For how m any polynomials does thereexist a polynomial of degree 3 such that ?SolutionSince has degree three, then has degree six. Thus, hasdegree six, so must have degree two, since has degree three.Hence, we conclude , , and must each be , , or . Since aquadratic is uniquely determined by three points, there can bedifferent quadratics after each of the values of , , and are chosen.However, we have included which are not quadratics. Nam ely,Clearly, we could not have included any other constant functions. For any linearfunction, we have . Again, it is pretty obvious that we have not included any other linear functions. Therefore, the desired answer is.Problem 25Let be the set of all points with coordinates , where and are eachchosen from the set . How many equilateral triangles have all their vertices in ?SolutionSolution 1 (non-rigorous)For this solution, we will just find as many solutions as possible, until it becom es intuitive that there are no m ore triangles left.Take an unit cube. We try to m ake three of its vertices form an equilateral triangle. This we find is possible by taking any vertex, and connecting the three adjacentvertices into a triangle. This triangle will have a side length of ; a quick further examination of this cube will show us that this is the only possible side length. Each of these triangles is determined by one vertex of the cube, so in one cube we have 8 equilateral triangles. We have 8 unit cubes, and then the entire cube, giving us 9cubes and equilateral triangles.It m ay be tem pting to connect the centers of the faces and to call that a cube, but a quick look at this tells us that that figure is actually an octahedron.Now, we look for any additional equilateral triangles. Since the space diagonals of the cube cannot m ake an equilateral triangle, we will assume symmetry in the cube.A bit m ore searching shows us that connecting the midpoints of three non-adjacent, non-parallel edges gives us m ore equilateral triangles.Notice that picking these three edges leaves two vertices alone, and that picking any two opposite vertices determine two equilateral triangles. Hence there areadditional equilateral triangles, for a total of .Solution 2 (rigorous)The three dim ensional distance form ula shows that the lengths of the equilateraltriangle must be , which yields the possible edge lengths ofSom e casework shows that are the only lengths that work, from which we can use the sam e counting argument as above.See Math Jam solution.。

合集下载

2005 AMC 10A Problems

2005 AMC 10A Problems

2005 AMC 10A ProblemsProblem 1While eating out, Mike and Joe each tipped their server 2 dollars. Mike tipped 10% of his bill and Joe tipped 20% of his bill. What was the difference, in dollars between their bills?（A ）2 （B ）4 （C ）5 （D ）10 （E ）20Problem 2For each pair of real numbers b a ≠, define the operation as.What is the value of? （A ）32- （B ）51- （C ）0 （D ）21 （E ）This value is not defined. Problem 3The equations 372=+x and 210-=-bx have the same solution. What is the value of ?（A ）-8 （B ）-4 （C ）2 （D ）4 （E ）8Problem 4A rectangle with a diagonal of length is twice as long as it is wide. What is the area ofthe rectangle?（A ）241x （B ）252x （C ）221x （D ）2x （E ）223x Problem 5A store normally sells windows at <dollar/>100 each. This week the store is offering one free window for each purchase of four. Dave needs seven windows and Doug needs eight windows. How many dollars will they save if they purchase the windows together rather than separately?（A ）100 （B ）200 （C ）300 （D ）400 （E ）500Problem 6The average (mean) of 20 numbers is 30, and the average of 30 other numbers is 20. What is the average of all 50 numbers?（A ）23 （B ）24 （C ）25 （D ）26 （E ）27Problem 7Josh and Mike live 13 miles apart. Yesterday Josh started to ride his bicycle toward Mike's house. A little later Mike started to ride his bicycle toward Josh's house. When they met, Josh had ridden for twice the length of time as Mike and at four-fifths of Mike's rate. How many miles had Mike ridden when they met?（A ）4 （B ）5 （C ）6 （D ）7 （E ）8Problem 8In the figure, the length of side AB of square ABCD is50 and BE =1. What is the areaof the inner square EFGH ?（A ）25 （B ）32 （C ）36 （D ）40 （E ）42Problem 9Three tiles are marked X and two other tiles are marked O . The five tiles are randomly arranged in a row. What is the probability that the arrangement reads XOXOX ?（A ）121 （B ）101 （C ）61 （D ）41 （E ）31 Problem 10 There are two values of for which the equation 09842=+++x ax x has only onesolution for . What is the sum of those values of ?（A ）-16 （B ）-8 （C ）0 （D ）8 （E ）20Problem 11A wooden cube units on a side is painted red on all six faces and then cut into 3n unitcubes. Exactly one-fourth of the total number of faces of the unit cubes are red. What is ?（A ）3 （B ）4 （C ）5 （D ）6 （E ）7 Problem 12The figure shown is called a trefoil and is constructed by drawing circular sectors about the sides of the congruent equilateral triangles. What is the area of a trefoil whose horizontal base has length 2?（A ）2331+π （B ）π32 （C ）4332+π （D ）3332+π （E ）2332+π Problem 13How many positive integers satisfy the following condition: 200100502)130(>>n n ?（A ）0 （B ）7 （C ）12 （D ）65 （E ）125Problem 14How many three-digit numbers satisfy the property that the middle digit is the average of the first and the last digits?（A ）41 （B ）42 （C ）43 （D ）44 （E ）45Problem 15How many positive cubes divide 3!·5!·7! ?（A ）2 （B ）3 （C ）4 （D ）5 （E ）6Problem 16The sum of the digits of a two-digit number is subtracted from the number. The units digit of the result is 6. How many two-digit numbers have this property?（A ）5 （B ）7 （C ）9 （D ）10 （E ）19Problem 17In the five-sided star shown, the letters A, B, C, D, and E are replaced by the numbers 3, 5, 6, 7 and 9, although not necessarily in this order. The sums of the numbers at the ends of the line segments AB, BC, CD, DE, and EA form an arithmetic sequence, although not necessarily in this order. What is the middle term of the sequence?（A ）9 （B ）10 （C ）11 （D ）12 （E ）13Problem 18Team A and team B play a series. The first team to win three games wins the series. Each team is equally likely to win each game, there are no ties, and the outcomes of the individual games are independent. If team B wins the second game and team A wins the series, what is the probability that team B wins the first game?（A ）51 （B ）41 （C ）31 （D ）21 （E ）32 Problem 19Three one-inch squares are placed with their bases on a line. The center square is lifted out and rotated 45 degrees, as shown. Then it is centered and lowered into its original location until it touches both of the adjoining squares. How many inches is the point from the line on which the bases of the original squares were placed?（A ）1 （B ）2 （C ）23 （D ）212 （E ）2 Problem 20 An equiangular octagon has four sides of length 1 and four sides of length 22, arrangedso that no two consecutive sides have the same length. What is the area of the octagon?（A ）27 （B ）227 （C ）2245+ （D ）2254+ （E ）7 Problem 21For how many positive integers does evenly divide ?（A ）3 （B ）5 （C ）7 （D ）9 （E ）11 Problem 22Let S be the set of the 2005 smallest positive multiples of 4, and let T be the set of the 2005 smallest positive multiples of 6. How many elements are common to S and T ?（A ）166 （B ）333 （C ）500 （D ）668 （E ）1001 Problem 23Let AB be a diameter of a circle and let C be a point on AB with 2·AC=BC . Let D and E be points on the circle such that DC ⊥AB and DE is a second diameter. What is the ratio of the area of △DCE to the area of △ABD ?（A ）61 （B ）41 （C ）31 （D ）21 （E ）32 Problem 24For each positive integer 1>m , let )(m P denote the greatest prime factor of . Forhow many positive integers is it true that both n n P =)( and48)48(+=+n n P ?（A ）0 （B ）1 （C ）3 （D ）4 （E ）5 Problem 25In ABC we have AB =25, BC =39, and AC =42. Points D and E are on AB and AC respectively, with AD =19 and AE =14. What is the ratio of the area of triangle ADE to the area of the quadrilateral BCED ? （A ）1521266 （B ）7519 （C ）31 （D ）5619 （E ）12005 AMC 10A SolutionsProblem 1Let be Mike's bill and be Joe's bill. 210010=m , so m =20; 210020=j ,so j =10So the desired difference is 101020=-=-j mProblem 2Problem 32372-=⇒=+x x , 42102-=⇒-=--b b , Problem 4Let the width of the rectangle be w .Then the length is 2w . Using the Pythagorean Theorem :5)2(222x w w w x =⇒+=, So the area of the rectangle is 2522x w w =∙ Problem 5The store's offer means that every 5th window is free.Dave would get free window. Doug would get free window.This is a total of 2 free windows. T ogether, they would getfree windows.So they get 3-2=1 additional window if they purchase the windows together.Therefore they save 1·100=100 Problem 6Since the average of the first 20 numbers is 30, their sum is 20*30=600.Since the average of 30 other numbers is 20, their sum is 30*20=600.So the sum of all 50 numbers is 600+600=1200Therefore, the average of all 50 numbers is 1200/50=24 Problem 7Let be the distance in miles that Mike rode.Since Josh rode for twice the length of time as Mike and at four-fifths of Mike's rate, he rode m m 58542=∙∙ miles.Since their combined distance was 13 miles, 51358=⇒=+m m m Problem 8(C) We see that side BE , which we know is 1, is also the shorter leg of one of the four right triangles (which are congruent, I'll not prove this). So, AH=1. Then HB=HE+BE=HE +1, and HE is one of the sides of the square whose area we want to find. So:650)1(1222=⇒=++HE HE , So, the area of the square is 3662=.Problem 9There are10!3!2!5= distinct arrangements of three 's and two 's. There is only 1 distinct arrangement that reads xoxoxTherefore the desired probability is 1/10 Problem 10A quadratic equation has exactly one root if and only if it is a perfect square . So set22)(984n mx x ax x +=+++, 22222984n mnx x m x ax x ++=+++Two polynomials are equal only if their coefficients are equal, so we must have242±=⇒=m m 392±=⇒=n n1232228±=∙∙±==+mn a , a =4 or a =-20So the desired sum is (4)+(-20)=-16Alternatively, note that whatever the two values of are, they must lead to equations ofthe form 02=++r qx px and 02=+-r qx px . So the two choices of must makeq a =+81 and q a -=+82 so 160162121-=+⇒=++a a a a Alternate SolutionSince this quadratic must have a double root, the discriminant of the quadratic formula for this quadratic must be 0. Therefore, we must have 144160)9)(4(4)8(22-+⇒=-+a a a . We can use the quadratic formula to solve for its roots (we can ignore the things in the radical sign as they will cancel out due to the sign when added). So we must have216216something something --++- Therefore, we have (-16)(2)/2=-16 Problem 11Since there are 2n little faces on each face of the big wooden cube , there are 26n little faces painted red. Since each unit cube has 6 faces, there are 36n little faces total.Since one-fourth of the little faces are painted red, 3266n n =441=⇒n Problem 12 The area of the trefoil is equal to the area of a small equilateral triangle plus the area of four 60ºsectors with a radius of 2/2=1 minus the area of a small equilateral triangle. This is equivalent to the area of four 60ºsectors with a radius of 1.So the answer is: ππ3213606042=∙∙∙ Problem 13We're given 200100502)130(>>n n , so 502005010050502)130(>>n n (because all termsare positive) and thus 422130>>n n . Solving each part seperatly:4162>⇒>n n , n n n >⇒>1301302, so 1304<<n . Therefore the answer is thenumber of positive integers over the interval (4, 130) which is 125. Problem 14Solution 1If the middle digit is the average of the first and last digits, twice the middle digit must be equal to the sum of the first and last digits.Doing some casework :If the middle digit is 1, possible numbers range from 111 to 210. So there are 2 numbers in this case.If the middle digit is 2, possible numbers range from 123 to 420. So there are 4 numbers in this case.If the middle digit is 3, possible numbers range from 135 to 630. So there are 6 numbers in this case.If the middle digit is 4, possible numbers range from 147 to 840. So there are 8 numbers in this case.If the middle digit is 5, possible numbers range from 159 to 951. So there are 9 numbers in this case.If the middle digit is 6, possible numbers range from 369 to 963. So there are 7 numbers in this case.If the middle digit is 7, possible numbers range from 579 to 975. So there are 5 numbers in this case.If the middle digit is 8, possible numbers range from 789 to 987. So there are 3 numbers in this case.If the middle digit is 9, the only possible number is 999. So there is number in this case.So the total number of three-digit numbers that satisfy the property is2+4+6+8+9+7+5+3+1=45 Solution 2 (much faster and slicker)Alternatively, we could note that the middle digit is uniquely defined by the first and third digits, since it is half of their sum. This also means that the sum of the first and third digits must be even. Since even numbers are formed either by adding two odd numbers or two even numbers, we can split our problem into 2 cases:If both the first digit and the last digit are odd, then we have 1, 3, 5, 7, or 9 as choices for each of these digits, and there are 5·5=25 numbers in this case.If both the first and last digits are even, then we have 2, 4, 6, 8 as our choices for the first digit and 0, 2, 4, 6, 8 for the third digit. There are 4·5=20 numbers here.The total number, then, is 20+25=45Problem 15Solution 13!·5!·7!=(3·2·1) ·(5·4·3·2·1) ·(7·6·5·4·3·2·1)=12487532∙∙∙ Therefore, a perfect cube that divides 3!·5!·7! must be in the form d c b a 7532∙∙∙where c b a ,,, and are nonnegativemultiples of 3 that are less than or equal to 8, 4, 2 and 1, respectively. So:( posibilities)( posibilities)( posibility) ( posibility)So the number of perfect cubes that divide 3!·5!·7! is 3·2·1·1=6 Solution 2If you factor 3!·5!·7! You get 247532∙∙. There are 3 ways for the first factor of acube:302,2, and 62. And the second ways are: 303,3. Answer : Problem 16Let the number be b a +10 where and are the tens and units digits of the number.So a b a b a 9)()10(=+-+ must have a units digit of 6This is only possible if 369=a , so 4=a is the only way this can be true.So the numbers that have this property are 40, 41, 42, 43, 44, 45, 46, 47, 48, 49.Therefore the answer is 10. Problem 17Each corner (a,b,c,d,e) goes to two sides/numbers. (A goes to AE and AB, D goes to DC and DE). The sum of every term is equal to 2(3+5+6+7+9)=60Since the middle term in an arithmetic sequence is the average of all the terms in the sequence, the middle number is 60/5=12 DProblem 18There are at most 5 games played.If team B won the first two games, team A would need to win the next three games. So the only possible order of wins is BBAAA.If team A won the first game, and team B won the second game, the possible order of wins are: ABBAA, ABABA, and ABAAX, where X denotes that the 5th game wasn't played.Since ABAAX is dependent on the outcome of games instead of 5, it is twice as likely to occur and can be treated as two possibilities.Since there is 1 possibility where team B wins the first game and 5 total possibilities, thedesired probability is 1/5.Problem 19Consider the rotated middle square shown in the figure. It will drop until lengthis 1 inch. Then, because DEC is a triangle, 22=EC , and 21=FC . Weknow that BC =2, so the distance from to the line is 2121+=+-FC BCProblem 20The area of the octagon can be divided up into 5 squares with side 22 and 4 right triangles, which are half the area of each of the squares. Therefore, the area of the octagon is equal to the area of 7)21(45=+squares. The area of each squareis 21)22(2=, so the area of 7 squares is 27 .Problem 21If evenly divides , then nn +++...216 is an integer . Since 2)1(...21+=+++n n n we may substitute the RHS in the above fraction . So the problem asks us for how many positive integers is 11262)1(+=+n n n n an integer, or equivalently when 12)1(=+n k for a positive integer .112+n is an integer when n +1 is a factor of 12. The factors of 12 are 1, 2, 3, 4, 6 and 12, so the possible values of are 0, 1, 2, 3, 5,and 11.But 0 isn't a positive integer, so only 1, 2, 3, 5, and 11 are possible values of . Thereforethe number of possible values of is 5Problem 22Since the least common multiple 12)6,4(1=cm , the elements that are common to S and T must be multiples of 12.Since 4·2005=8020 and 6·2005=12030, several multiples of 12 that are in T won't be in S , but all multiples of 12 that are in S will be in T . So we just need to find the number of multiples of 12 that are in S . Since 4·3=12 every 3rd element of S will be a multiple of 12Therefore the answer isProblem 23Solution 1Let us assume that the diameter is of length 1. AC is 31 of diameter and CO is 613121=-. OD is the radius of the circle, so using the Pythagorean theorem height CD of △AOCIs 32)61()21(22=-. This is also the height of the .Area of theis 362326121=∙∙. The height of can be found using the area ofand DO as base.Hence the height of is922121362=∙ . The diameter is the base for both thetrianglesand .Hence, the ratio of the area of to the area of is 313292=Solution 2Since and share a base, the ratio of their areas is the ratio of theiraltitudes. Draw the altitude from C to DE .r OC r OD 31,==. Since ︒=∠=∠90DFC m DCO m ,then . So the ratio of the two altitudes is31==DO OC DC CFProblem 24 If ,)(n n P =then 21p n =, where is a prime number .If 48)48(+=+n n P , then 2248p n =+ , where is a different prime number.So: 4822+=n p , ,21n p = 48))((4812122122=-+⇒=-p p p p p pSince )()(:011121p p p p p ->+> Looking at pairs of divisors of 48, we have several possibilities to solve for and :48)(12=+p p 1)(12=-p p 2/471=p 2/492=p24)(12=+p p 2)(12=-p p 111=p 132=p16)(12=+p p 3)(12=-p p 2/131=p 2/192=p12)(12=+p p 4)(12=-p p 41=p 82=p8)(12=+p p 6)(12=-p p 11=p 72=pThe only solution ),(21p p where both numbers are primes is (11,13).Therefore the number of positive integers that satisfy both statements is 1. Problem 25The area of a triangle is A bc sin 21. Using this formula: A A ADE sin 133sin 141921][=∙∙∙= A A ABC sin 525sin 422521][=∙∙∙= Since the area of BCED is equal to the area of ABC minus the area of ADE , A A A BCED sin 392sin 133sin 525][=-=Therefore, thedesired ratio is sin 392sin 133=A A 2005 AMC 10A Answer Key1. D2. C3. B4. B5. A6.B7.B8.C 8. B 10. A11.B 12. B 13.E 14.E 15. E 16.D 17. D 18.A 19.D 20.A21.B 22.D 23.C 24.B 25.D。

2005年12月A级试题及答案解析

2005年12月A级试题及答案解析

2005年12月A级试题及答案解析2005年12月高等学校英语应用能力A级试题Part I Listening Comprehension (15 minutes)Directions: This part is to test your listening ability. It consists of 3 sections.Section ADirections: This section is to test your ability to understand short dialogues. There are 5 recorded dialogues in it.After each dialogue, there is a recorded question. Both the dialogues and questions will be spokenonly once. When you hear a question, you should decide on the correct answer from the 4 choicesmarked A, B, C and D given in your test paper. Then you should mark the corresponding letter onthe Answer Sheet with a single line through the center.1. A. In a post office. B. In a bank. C. In a big store. D. In a hotel.2. A. He has a lot of experience. B. He appears to be a bit too quiet.C. He is nice-looking.D. He lacks experience.3. A. Prepare a price list. B. Go to the Brown Company.C. Send a parcel by post.D. Mail a price list.4. A. She should repair the telephone herself. B. She can turn to Mike for help.C. He knows nothing about telephone.D. He will call the telephone company.5. A. Seeing a film. B. Attending an evening party.C. Having another try.D. Finding a better theater.Section BDirections: This section is to test your ability to understand short conversations. There are 2 recorded conversations in it. After each conversation, there are some recorded questions. Both theconversations and questions will be spoken two times. When you hear a question, you should decideon the correct answer from the 4 choices marked A, B, C and D given in your test paper. Then youshould mark the corresponding letter on the Answer Sheet with a single line through the center. Conversation 16. A. In the office. B. In a factory.C. At the airport.D. At the railway station.7. A. She was busy working. B. She was having a holiday.C. She wanted to get relaxed.D. She wanted to shop for Christmas.Conversation 28. A. To make an apology. B. To place an order.C. To ask about delivery.D. To cancel an appointment.9. A. Trucks B. Printers C. Phones D. Cameras10. A. For the late payment. B. For the damaged products.C. For the lost goods.D. For the delayed delivery.Section CDirections: This section is to test your ability to comprehend short passages. Y ou will hear a recorded passage.After that you will hear five questions. Both the passage and the questions will be read two times.When you hear a question, you should complete the answer to it with a word or a short phrase (in notmore than 3 words). The questions and incomplete answers are printed in your test paper. Y ou shouldwrite your answer on the Answer Sheet correspondingly. Now the passage will begin.11. When did the family car become common in America?In the early .12. Why do more Americans have to drive to work?Because they have moved out of the to the suburbs.13. Why do people like smaller cars now?Because the cost of gas has .14. How popular are Japanese and German cars in the U.S.?They sell .15. What does the speaker mainly talk about?in the U.S.Part II Structure (15 minutes)Directions: This part is to test your ability to construct grammatically correct sentences.It consists of 2 sections. Section ADirections: There are 10 incomplete statements here. Y ou are required to complete each statement by choosing the appropriate answer from the 4 choices marked A, B, C and D. Y ou should mark the corresponding letter on the Answer Sheet with a single line through the center.16. My boss said that he was badly need of my assistance.A. atB. inD. with17. She didn’t know to express her ideas clearly when she was invited to speak at a meeting.A. whereB. whyC. whatD. how18. The professor, as a splendid speaker, was warmly received by the students.A. knownB. knowingC. to be knownD. having known19. tired of Tom’s all-talk-no-action attitude, Julia decided to do the job all by herself.A. To getB. To have gotC. GettingD. Have got20. He in this company since he graduated from Andong Technical College ten years ago.A. workedB. has been workingC. had workedD. was working21. you have any questions or needs, please contact the manager after 5:00 p.m. on weekdays.A. BecauseB. WhereD. Though22. It cost her a lot of money, but she doesn’t regret a year traveling around the world.A. to have spentB. to spendC. spentD. spending23. After for the job, you will be required to take a language test.A. being interviewedB. being interviewingC. interviewingD. having interviewed24. He is used to flying by air and on no occasion frightened.A. he has ever feltB. he ever feelsC. ever does he feelD. has he ever felt25. It is most desirable that he for the information by himself with a few clicks online.A. searchB. searchedC. has searchedD. will searchSection BDirections: There are also 10 incomplete statements here. Y ou should fill in each blank with the proper form of the word given in the brackets. Write the word or words in the corresponding space on the Answer26. Follow the (settle) of the strike, the train service is now back to normal.27. By the end of this year the factory (produce) 20,000 cell phones.28. Those who are working in the Human Resources Department are (humorous ) than we expected.29. We should read more and see more in order to (wide) our horizons.30. All the members of the club were present when the Chairman (elect) last week.31. When she was very young, my sister already knew where (put) her toys and dolls.32. Some experts suggest that we slow down the (economy) growth in the country.33. He might have been killed in that car accident yesterday if he (take) part in that activity with theteam.34. Doing a second job to earn more money also means you have to pay (addition)income tax.35. (watch) by a crowd of people, Charles felt embarrassed and couldn’t say a word.Part III Reading Comprehension (40 minutes)Directions: This part is to test your reading ability. There are 5 tasks for you to fulfill. Y ou should read the material carefully and do the tasks as you are instructed.T ask 1Directions: After reading the following passage, you will find 5 questions or unfinished statements, marked 36 to40. For each question or statement there are 4 choices marked A, B, C and D. Y ou should make thecorrect choice and mark the corresponding letter on the Answer Sheet with a single line through thecenter.The holiday shopping season got off to a flying start over the Thanksgiving (感恩节) weekend in the U.S. Retailers (零售商) attracted discount-hungry consumers with specials (特价商品) of televisions, DVD players and other gifts. Discounters like Wal-Mart and Target had the greatest success, offering customers early bird specials, while major department stores and clothing chains only met modest sales goals. Online sales were also strong.“It was as expected. It wasn’t a source of great prosperity, but it was a good start,”said Nevill, a man in a US reta il consulting firm. He said, “One thing we often heard was that only things marked down (降价) were selling. Consumers want a bargain, especially around the holiday.”Shoppers did seem to like the many markdowns. A lady in the State of Connecticut said, “I’ve never seen so many sales. I feel like a kid in a candy store.”Nevill estimated that discounters had a 4% gain over last year’s Thanksgiving figure, while sales at department stores and mall based clothing chains were up 2% for the weekend. However, the Thanksgiving weekend is not necessarily a good way to judge how retailers will be successful for the season. Over the past few years, the weekend accounted for less than 10% of sales.Online merchants also had a good weekend. Sales were up 61% last Friday, compared to the day after Thanksgiving a year ago.36. Over the Thanksgiving weekend, people bought a lot more than usual because .A. Wal-Mart and T arget offered birds as giftsB. shopping was the major weekend activityC. people had time to go shopping onlineD. they were allowed special discounts37. Retailers have the experience that consumers seem to .A. feel eager to buy discounted goodsB. be hungry for special kinds of goodsC. like to buy new DVD players as giftsD. be interested in major department stores38. From the context we know that Wal-Mart and Target are the names of two .A. food storesB. large retailersC. online servicesD. clothing chains39. By saying “I feel like a kid in a candy store”, the lady in Connecticut probably means that .A. she has difficulty choosing form such a wide variety of goodsB. she cannot decide whether she should buy some candyC. she is quite interested in buying candyD. she recalls her happy childhood days40. Which of the following statements is TRUE according to the passage?A. Online sales were 61% higher than retailers’.B. Online merchants had an average weekend sale.C. Retailers lost much profit because of the discounts they offered.D. Retailers had an increase in sales in the Thanksgiving weekend.T ask 2Directions: This task is the same as Task 1. The 5 questions or unfinished statements are numbered 41 to 45.The Future of General Motors (GM) Is Y oursWe have everything college students need to know about GM’s Cooperative Education and Inter n (实习) Programs in our Student Center. Each fall, GM recruiters (招聘人员) visit the campuses of many of the nation’s top engineering and business colleges and universities to recruit students. These students are considered for interesting assignments throughout our U.S. operations. Check out our Recruiting Calendar to see if GM will be visiting your campus. If your campus is not listed, please apply online.Will GM start your career moving? Fasten your seat belt!For full-time college students, General Motors offers both a Cooperative Education and an Intern Program. Participants in these real-business-world educational programs gain valuable degree-related experience, develop an insider’s understanding of how GM works and earn competitive wages. These programs are designed to provide GM with a source of highly talented candidates while giving students an opportunity for hands-on experience in their chosen field. As a result, participants in these programs are given serious consideration for full-time positions with GM when they graduate. Candidates for these programs must successfully complete an online assessment and possess qualifications that match the business needs of the organization.41. GM is likely to recruit college students for its programs who .A. do a full-time college programB. major in engineering and businessC. have had some practical experienceD. have gathered information about GM42. The Recruiting Calendar (Line 4, Para.1) gives the information about .A. the kinds of people GM needs to trainB. the nation’s top colleges and universitiesC. GM recruiters’ visits to colleges and universities.D. The interesting tasks GM expects the students to fulfill43. Those students whose university is not listed on the Recruiting Calendar may .A. apply to GM online.B. be given interesting tasksC. come to the GM’s training offices directlyD. invite recruiters to visit their universities.44. The programs that GM offers to full-time college students will help them to .A. gain information about business in generalB. get business experience and good wagesC. develop their talents fullyD. obtain a higher degree45. GM offers the Cooperative Education and Intern Program in order to .A. make its business needs known to the publicB. perform successful online assessmentsC. advertise its newly-designed productsD. find out highly talented candidatesT ask 3Directions:The following is an Introduction to Montserrat Publishing. After reading it, you are required to complete the outline below it (No.46 to No.50).Y ou should write your answerbriefly (in not morethan 3 words) on the Answer Sheet correspondingly.We at Montserrat Publishing have published two books devoted to improving the everyday vocabulary of intermediate and advanced English language students.Y ou probably find that even though you have been studying English for many years, you still have problems following a social conversation between English speaking people.Maybe you need English for business and you have a wide knowledge of the vocabulary related to your specialty (专业), but you find it difficult to conduct meetings in English. This is because you cannot understand the practical, everyday expressions being employed. Y ou may also need to understand journals and trade magazines written in English where a dictionary is not enough to convey the real meaning of what is being said.The Practical Everyday English series consists of a collection of words and expressions used every day by all classes of society. The idea is that after being presented with a word or expression the reader can see how it is used by reading two or more clear examples. Each example contains at least one word or even two or three which have been studied on earlier pages.Information on Newly-published BooksName of the series: (46)Number of the books: (47)Intended readers:(48) students studying EnglishPurposes: 1. to enlarge the students’ everyday (49) of English2. to help understand the meaning of practical, everyday expressionsContent: a collection of (50) used every day in varioussituationsT ask 4Directions: The following is a list of terms from a book on WTO. After reading it, you are required to find the items equivalent to (与……等同) those given in Chinese in the table below.Then you should put thecorresponding letters in brackets on the Answer sheet, numbered 51 through 55.A—dispute settlement body B—balance of international paymentsC— world trade organization D— risk managementE—investment in non-productive project F—grant the national treatmentG— appeal body H— common agriculture policyI— customs values J— export performanceK—food security L—free riderM— grey area measures N— import licensingO— market access P— market boardsQ—presence of national parson R—north american free trade areaS— international settlement T— peace clauseU—least-developed countries V—most-favored-nation treatmentExample:(F)实行国民待遇(P)营销机构51. ( )上诉机构( )最惠国待遇52. ( )和平条款( )进口许可53. ( )市场准入( )非生产性投资54. ( )食品安全保障( )共同农业政策55. ( )风险管理( )争端解决机构T ask 5Directions: There is an advertisement below. After reading it, you are required to complete the statements that follow the questions (No.56 to No.60). Y ou should write your answers (in not more than 3 words) onthe Answer Sheet correspondingly.Douglass Company (hereafter〈以下〉referred to as the Buyer) as one party and Anhui Import and Export Corporation (hereafter referred to as the Seller) as the other party agree to sign by their authorized representatives, as a result of friendly negotiation, the present Contract under the following terms and conditions:1) The Seller will arrange and sell 260 sets of drilling machines in the following two fiscal years (财政年度);2) The Seller should take care of the quality, quantity, packing, and shipping;3) The Buyer should pay the Seller by irrevocable L/C and take care of the insurance;4) Claim shall be made within 10 days of goods’ arrival;5) Neither party shall break the contract without mutual agreement;6) The present contract is written in English, valid for two years, after which it may be extended, amended ordiscontinued;7) The contract was signed on April 14, 2005 in Hefei by representatives of the two parties.(Signature) For the Anhui Mechanic Import and Export Co. of People’s Republic of China.(Signature) For the Douglass Company of New Y ork, N.Y, U.S.A.56. What is the contract product?.57. Which party is responsible for the insurance?.58. How should the Buyer pay the seller?By .59. How long is the contract valid for?.60. Who signed the contract?The of the two parties.Part IV T ranslation — English into Chinese (25 minutes)Directions: This part is to test your ability to translate English into Chinese. Each of the four sentences (No.61 to No.64) is followed by four choices of suggested Chinese translation marked A, B, C and D. Markthe best choice and write the corresponding letter on the Answer Sheet. Write your translation of theparagraph (No.65) in the corresponding space on the Translation / Composition Sheet.61. Y ou can buy a three-year-old car for only 60% of the price for a new vehicle and still have several years oftrouble-free driving.A. 你可以买一辆已使用3年的二手车，价格只是新车的60％，但你仍然可以开几年而不出故障。

美国数学竞赛AMC8 -- 2005年真题解析(英文解析+中文解析)

美国数学竞赛AMC8 -- 2005年真题解析(英文解析+中文解析)

美国数学竞赛AMC8 – 2005年真题解析(英文解析+中文解析)Problem 1Answer: BSolution:If x is the number, then 2x=60 and x=30. Dividing the number by 2 yields 15.中文解析：按照Connie的计算，这个数乘以2是60，可知这个数是30. 应该做的计算是30除以2，因而正确答案应该是15. 答案是B。

Problem 2Answer: CSolution:Karl paid 5*2.5=$12.5. 20% of this cost that he saved is 12.5*0.2=$2.5.中文解析：Karl按原价买了5个文件夹，支付的费用是：2.5*5=12.5. 折扣价是：1.25*0.8=10。

如果Karl 等一天，可以省2.5元。

答案是C.Problem 3Answer: DSolution:Rotating square ABCD counterclockwise 45° so that the line of symmetry BD is a vertical line makes it easier to see that 4 squares need to be colored to match its corresponding square.中文解析：如上图所示，以BD为对称轴，标蓝色的方块需要涂黑。

共4块，答案是D。

Problem 4Answer: CSolution:The perimeter of the triangle is 6.1+8.2+9.7=24cm. A square's perimeter is four times its side length, since all its side lengths are equal. If the square's perimeter is 24, the side length is24/4=6, and the area is 6*6=36.中文解析：三角形的周长是：6.1+8.2+9.7=24. 正方形的周长和三角形相等，也是24，则其边长是24/4=6. 其面积是：6*6=36. 答案是C。

2005AMC12A

2005AMC12A

1.Two is 10%of x and 20%of y .What is x −y ?(A)1(B)2(C)5(D)10(E)202.The equations 2x +7=3and bx −10=−2have the same solution for x .What is the value of b ?(A)−8(B)−4(C)−2(D)4(E)83.A rectangle with a diagonal of length x is twice as long as it is wide.What is the area of the rectangle?(A)14x 2(B)25x 2(C)12x 2(D)x 2(E)32x 24.A store normally sells windows at $100each.This week the store is offering one free window for each purchase of four.Dave needs seven windows and Doug needs eight windows.How many dollars will they save if they purchase the windows together rather than separately?(A)100(B)200(C)300(D)400(E)5005.The average (mean)of 20numbers is 30,and the average of 30other numbers is 20.What is the average of all 50numbers?(A)23(B)24(C)25(D)26(E)276.Josh and Mike live 13miles apart.Yesterday Josh started to ride his bicycle toward Mike’s house.A little later Mike started to ride his bicycle toward Josh’s house.When they met,Josh had ridden for twice the length of time as Mike and at four-fifths of Mike’s rate.How many miles had Mike ridden when they met?(A)4(B)5(C)6(D)7(E)87.Square EF GH is inside square ABCD so that each side of EF GH can be extended to pass through a vertex of ABCD .Square ABCD has side length √50,E is between B and H ,and BE =1.What is the area of the inner square EF GH ?A BC DGHF E(A)25(B)32(C)36(D)40(E)428.Let A,M,and C be digits with (100A +10M +C )(A +M +C )=2005.What is A ?(A)1(B)2(C)3(D)4(E)59.There are two values of a for which the equation 4x 2+ax +8x +9=0has only one solution for x .What is the sum of those values of a ?(A)−16(B)−8(C)0(D)8(E)2010.A wooden cube n units on a side is painted red on all six faces and then cutinto n 3unit cubes.Exactly one-fourth of the total number of faces of the unit cubes are red.What is n ?(A)3(B)4(C)5(D)6(E)711.How many three-digit numbers satisfy the property that the middle digit is theaverage of the first and the last digits?(A)41(B)42(C)43(D)44(E)4512.A line passes through A (1,1)and B (100,1000).How many other points withinteger coordinates are on the line and strictly between A and B ?(A)0(B)2(C)3(D)8(E)913.In the five-sided star shown,the letters A ,B ,C ,D and E are replaced by thenumbers 3,5,6,7and 9,although not necessarily in that order.The sums of the numbers at the ends of the line segments AB ,BC ,CD ,DE and EA form an arithmetic sequence,although not necessarily in that order.What is the middle term of the arithmetic sequence?AB C DE(A)9(B)10(C)11(D)12(E)1314.On a standard die one of the dots is removed at random with each dot equallylikely to be chosen.The die is then rolled.What is the probability that the top face has an odd number of dots?(A)511(B)1021(C)12(D)1121(E)61115.Let AB be a diameter of a circle and C be a point on AB with2·AC=BC.Let D and E be points on the circle such that DC⊥AB and DE is a second diameter.What is the ratio of the area of DCE to the area of ABD?B(A)16(B)14(C)13(D)12(E)2316.Three circles of radius s are drawn in thefirst quadrant of the xy-plane.Thefirst circle is tangent to both axes,the second is tangent to thefirst circle and the x-axis,and the third is tangent to thefirst circle and the y-axis.A circle of radius r>s is tangent to both axes and to the second and third circles.What is r/s?(A)5(B)6(C)8(D)9(E)1017.A unit cube is cut twice to form three triangular prisms,two of which arecongruent,as shown in Figure1.The cube is then cut in the same manner along the dashed lines shown in Figure2.This creates nine pieces.What is the volume of the piece that contains vertex W?(A)112(B)19(C)18(D)16(E)1418.Call a number“prime-looking”if it is composite but not divisible by2,3,or5.The three smallest prime-looking numbers are49,77,and91.There are168prime numbers less than1000.How many prime-looking numbers are there less than1000?(A)100(B)102(C)104(D)106(E)10819.A faulty car odometer proceeds from digit3to digit5,always skipping the digit4,regardless of position.For example,after traveling one mile the odometer changed from000039to000050.If the odometer now reads002005,how many miles has the car actually traveled?(A)1404(B)1462(C)1604(D)1605(E)180420.For each x in[0,1],definef(x)=2x,if0≤x≤12 2−2x,if12<x≤1.Let f[2](x)=f(f(x)),and f[n+1](x)=f[n](f(x))for each integer n≥2.For how many values of x in[0,1]is f[2005](x)=1/2?(A)0(B)2005(C)4010(D)20052(E)2200521.How many ordered triples of integers(a,b,c),with a≥2,b≥1,and c≥0,satisfy both log a b=c2005and a+b+c=2005?(A)0(B)1(C)2(D)3(E)422.A rectangular box P is inscribed in a sphere of radius r.The surface area of Pis384,and the sum of the lengths of its12edges is112.What is r?(A)8(B)10(C)12(D)14(E)1623.Two distinct numbers a and b are chosen randomly from the set{2,22,23,...,225}.What is the probability that log a b is an integer?(A)225(B)31300(C)13100(D)750(E)1224.Let P(x)=(x−1)(x−2)(x−3).For how many polynomials Q(x)does thereexist a polynomial R(x)of degree3such that P(Q(x))=P(x)·R(x)?(A)19(B)22(C)24(D)27(E)3225.Let S be the set of all points with coordinates(x,y,z),where x,y,and z areeach chosen from the set{0,1,2}.How many equilateral triangles have all their vertices in S?(A)72(B)76(C)80(D)84(E)88。